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प्रश्न
\[\tan x + \tan\left( \frac{\pi}{3} + x \right) - \tan\left( \frac{\pi}{3} - x \right) = 3 \tan 3x\]
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उत्तर
\[\frac{\pi}{3} = 60°\]
\[LHS = \text{ tan } x + \tan\left( 60° + x \right) - \tan\left( 60° - x \right)\]
\[ = \text{ tan } x + \left( \frac{\tan60° + \text { tan } x}{1 - \tan60° \text{ tan } x} \right) - \left( \frac{\tan60° - \text{ tan } x}{1 + \tan60° \text{ tan } x} \right)\]
\[ \left[ \tan\left( x + y \right) = \frac{\text{ tan } x + \text{ tan } y}{1 - \text{ tan } x \text{ tan } y} \text{ and } \tan\left( x - y \right) = \frac{\text{ tan } x - \text{ tan } y}{1 + \text{ tan } x \text{ tan } y} \right]\]
\[ = \text{ tanx } + \frac{\sqrt{3} + \text{ tan } x}{1 - \sqrt{3} \text{ tan } x} - \frac{\sqrt{3} - \text{ tan } x}{1 + \sqrt{3} \text{ tan } x}\]
\[ = \text{ tan } x + \frac{\sqrt{3} + 3\text{ tan } x + \text{ tan } x + \sqrt{3} \tan^2 x + \sqrt{3} + 3\text{ tan } x + \text{ tan } x - \sqrt{3} \tan^2 x}{\left( 1 - \sqrt{3} \text{ tan } x \right)\left( 1 + \sqrt{3} \text{ tan } x \right)}\]
\[ = \text{ tan } x + \frac{8\text{ tan } x}{1 - 3 \tan^2 x}\]
\[= \frac{\text{ tan } x - 3 \tan^3 x + 8\text{ tan } x}{1 - 3 \tan^2 x}\]
\[ = \frac{9\text{ tan } x - 3 \tan^3 x}{1 - 3 \tan^2 x}\]
\[ = 3\left( \frac{3\text{ tan } x - \tan^3 x}{1 - 3 \tan^2 x} \right) \left( \because \tan3\theta = \frac{3tan\theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) \]
\[ = 3\tan3x\]
\[ = RHS\]
\[\text{ Hence proved .} \]
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